91Ó°ÊÓ

Now let's consider the function \(g(x) = x\). This is a simple linear function with slope \(1\) and no vertical shift. It passes through the origin \((0, 0)\) and increases with \(x\). #Step 3: Analyze intersection points between f(x) and g(x)#

Now we need to show that the intersection point between \(f(x)\) and \(g(x)\) has positive coordinates and is within the range of the problem statement. We know that \(0 < a < 1\) and \(b > 0\), which means the minimum value of \(f(x)\) is \(b > 0\). This means the entire graph of \(f(x)\) is above the \(x\)-axis, and since \(g(x)\) passes through the origin and increases with \(x\), the intersection point \((x_1, y_1)\) must have positive coordinates. Moreover, note that the maximum value of \(f(x)\) is obtained when \(\sin x = 1\). In this case, the maximum value, \(f_{max}\), is given by: \[f_{max} = a\sin x + b = a + b\] Since \(g(x)\) increases with \(x\) and \(f(x)\) decreases after achieving the maximum value, the intersection point between \(g(x)\) and \(f(x)\) must have \(x \leq a + b\). Thus, we have shown that the equation \(x = a\sin x + b\) has at least one positive root \(x\) which does not exceed \(a + b\).